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RMS & Avg. of waves

  1. May 25, 2009 #1
    1. The problem statement, all variables and given/known data

    I need to find the average and RMS voltages for the waves (See attached pictures) but I'm not sure exactly how to do it as I don't know how to get an equation for V(t)


    2. Relevant equations
    I Know:

    VRMS = [tex]\sqrt{\frac{1}{T}\int V(t)^2 dt}[/tex] between 0 and T

    And for Avg it's just VAVG = [tex] \frac{1}{T}\int V(t) dt[/tex] between 0 and T


    3. The attempt at a solution


    I thought I could find the averages by finding the total area for 1 period, then dividing by 1 period. This works (I think, as I don't have the answers)

    Giving for the Triangular wave: [tex]\frac{T}{3} \times \frac{10}{T} = 3.3V[/tex]

    and for the second sawtooth type wave: [tex] \frac{\frac{1}{2}\frac{T}{3} \times 1}{T} = \frac{1}{6}V[/tex]


    Thanks
     

    Attached Files:

  2. jcsd
  3. May 25, 2009 #2

    Redbelly98

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    Looks good for VAVG.

    For VRMS, you'll need to write equations for the waveforms. To do that, you may choose any point to represent t=0.
     
  4. May 25, 2009 #3
    That's the problem, I don't know what to write the equations as. The triangular wave has 3 functions. One for 0 - [tex]\frac{T}{3}[/tex] (upwards slanting triangle) one for 0 - [tex]\frac{2T}{3} [/tex] (downwards slanting triangle) and one for the last part of the waveform, which is just V(t) = 0.

    What are the waveforms for the first 2 sections? And can they be summed like so?:

    VRMS = [tex]\sqrt{\frac{1}{T}\int V_A(t)^2 dt} + \sqrt{\frac{1}{T}\int V_B(t)^2 dt}[/tex] the first integral being from 0 to T/3 and the second from T/3 to 2T/3
     
  5. May 25, 2009 #4

    Redbelly98

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    That's pretty much the idea, except that it would be

    [tex]
    \sqrt{\frac{1}{T}\int V_A(t)^2 dt \ + \ \frac{1}{T}\int V_B(t)^2 dt}
    [/tex]

    with, as you said, "the first integral being from 0 to T/3 and the second from T/3 to 2T/3"
     
  6. May 25, 2009 #5
    I still can't seem to figure out the RMS values. The problem I have is determining the functions V(t) for each section of the wave.

    I have an exam on this tomorrow so any help would be appreciated!
     
  7. May 25, 2009 #6

    Redbelly98

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    Note that the function is composed of straight line segments. Do you remember, from high school algebra,

    y = mx + b

    where m is the slope and b is the y-intercept?

    For example . . .

    . . . for the triangular wave, and 0 ≤ t ≤ T/3:

    v(t) = m t + b

    What is the slope m in this interval (0 ≤ t ≤ T/3)?
    What is the intercept b in this interval?

    Find m and b in this interval, and you'll have the function.

    Do the same for the interval T/3 ≤ t ≤ 2T/3, and you'll have v(t) in that interval as well.
     
  8. May 25, 2009 #7
    Hmm...
    using v(t) = m t + b
    I get:
    [tex]V_A(t) = \frac{30t}{T}[/tex]
    [tex]V_B(t) = 20 - \frac{30t}{T}[/tex]

    So I tried integrating all of this, then taking the root (long and tiring process!) and I got V_RMS = 4.71V

    Does this sound about right?

    Here's the integration I did (without the square root)

    [tex]\sqrt{\frac{30^2}{T^3}\int t^2 dt + \frac{1}{T}\int 400t dt - \frac{1}{T^2}\int 600t^2 dt + \frac{1}{T^3}\int 300t^3 dt}[/tex]

    With the first being between 0 and T/3 and the last 3 parts being between T/3 and 2T/3
    (there's 3 parts because of squaring [tex]20 - \frac{30t}{T}[/tex])

    I hope this is right!
     
  9. May 25, 2009 #8

    Redbelly98

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    You're on the right track, but I do see a couple of errors.

    You seem to have expanded (20 - 30t/T)2 incorrectly, in particular the "middle" term's coefficient of -600 is wrong.

    Also, there is an extra factor of t in 3 of your integrals.
     
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